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I am trying to prove the following: the ring $$R=\frac{\mathbb{Z}[x_1,x_2,x_3,\cdots]}{(x_1-x_2^2,x_2-x_3^2,\cdots)}$$does not have the ascending chain condition for principal ideals.

By construction, we have $x_i=x_{i+1}^2$ in $R$. So there is an obvious candidate for an infinite chain of principal ideals, namely, $$(x_1)\subset(x_2)\subset\cdots.$$ However, I have been unable to show that the inclusions in this chain (or at least infinitely many of them) are strict. I have managed to prove it (see below), but I am unsure if my argument is correct. So please tell me

  • if my argument is valid, and
  • if there is a better way to prove this.

Any help is appreciated. Thanks in advance!


My attempt: Consider the ring homomorphism $$\varphi:\mathbb{Z}[x_1,x_2,\cdots]\to\mathbb R$$ which extends the inclusion $\mathbb{Z}\to\mathbb R$ and maps $x_i\to 2^{2^{-n}}$. Because $\left(2^{2^{-(n+1)}}\ \ \right)^2=2^{2^{-n}}$, the kernel of this homomrphism contains the ideal $(x_1-x_2^2,x_2-x_3^2,\cdots)$. So it descends to a surjective ring homomorphism $\overline{\varphi}:R\to\operatorname{im}\varphi$.

Now suppose that $(x_1)=(x_2)$ in $R$. Since $\overline{\varphi}((x_n))=(2^{2^{-n}})$, this means that $(2^{2^{-1}})=(2^{2^{-2}})$ in $\operatorname{im}\varphi$. But $\operatorname{im}\varphi$ is an integral domain and its units are $\pm 1$, so $2^{2^{-1}}=\pm 2^{2^{-2}}$, a contradiction.

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There's a big gap in your proof: you've never shown that the units of $\operatorname{im}\varphi$ are just $\pm 1$. In fact, this is false: for instance, $(\sqrt{2}+1)(\sqrt{2}-1)=1$ so $\sqrt{2}\pm 1$ are units in $\operatorname{im}\varphi$.

Here's how I would show it. Suppose one of the inclusions is not strict, i.e. that $x_i$ divides $x_{i+1}$ for some $i$. The fact that $x_i$ divides $x_{i+1}$ is witnessed by finitely many of the generators and relations of $R$: there is a polynomial $p\in \mathbb{Z}[x_1,x_2,\dots]$ such that $x_ip-x_{i+1}\in (x_1-x_2^2,x_2-x_3^2)\dots$, and then there are only finitely many variables appearing in $p$ and finitely many variables needed to write $x_ip-x_{i+1}$ as a linear combination of $x_1-x_2^2,x_2-x_3^2,\dots$. This means that there is some $n$ such that $x_i$ divides $x_{i+1}$ in the ring $R_n=\mathbb{Z}[x_1,\dots,x_n]/(x_1-x_2^2,x_2-x_3^3\dots,x_{n-1}-x_n^2)$ which has only the generators and relations of $R$ up to $x_n$. But $R_n$ is easily seen to be isomorphic to just $\mathbb{Z}[x]$, by mapping $x_j$ to $x^{2^{n-j}}$. So to say that $x_i$ divides $x_{i+1}$ would mean that $x^{2^{n-i}}$ divides $x^{2^{n-i-1}}$ in $\mathbb{Z}[x]$ which is false.

Alternatively, for an approach that is closer to yours, you could define a ring $\mathbb{Z}[x,x^{1/2},x^{1/4},\dots]$ of formal linear combinations of nonnegative dyadic rational powers of $x$ (i.e., the monoid ring of the monoid of nonnegative dyadic rational numbers). You can then define a homomorphism $\varphi:R\to\mathbb{Z}[x,x^{1/2},x^{1/4},\dots]$ sending $x_n$ to $x^{2^{-n}}$ (in fact, this $\varphi$ is an isomorphism). If $x_i$ divided $x_{i+1}$, we would conclude that $x^{2^{-i}}$ divides $x^{2^{-i-1}}$ and so they differ by a unit. But the only units of $\mathbb{Z}[x,x^{1/2},x^{1/4},\dots]$ are $\pm 1$ (because degrees add when you multiply, just like with ordinary polynomials), and so this is a contradiction.

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